$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{Y$Y$ is a non-negative random variable, not necessary integrable. How to show
$$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{Y<n}Y \, dP=0$$
My idea is to find a good upper bound that converges to zero. However, I didn't find a good one. My upper bound is 
$$\frac{1}{n}\int_{Y<n}Y \, dP\leq \frac{1}{n}\int_{Y<n}n \, dP\leq \int_{Y<n}dP\leqslant 1.$$
 A: Note that the sequence of random variables $X_n = \frac{1}{n} I\{Y < n\} Y$ is bounded by $1$ and converges pointwise to $0$. The assertion now follows from the dominated convergence theorem.
A: Here is a direct proof (which doesn't require any additional theorems such as the dominated convergence theorem):
Fix $\epsilon>0$. Since $A_n := \{n \leq Y < \infty\}$ is a sequence of decreasing sets satisfying $\bigcap_{n \in \mathbb{N}} A_n = \emptyset$, the continuity of the (probability) measure $\mathbb{P}$ gives
$$\lim_{n \to \infty} \mathbb{P}(n \leq Y < \infty) = 0;$$
in particular we can choose $N \in \mathbb{N}$ such that $\mathbb{P}(N \leq Y < \infty) \leq \epsilon$. Now
$$\begin{align*} \frac{1}{n} \int_{Y<n} Y \, d\mathbb{P} &= \frac{1}{n} \int_{Y<N} Y \, d\mathbb{P} + \int_{N \leq Y < n} \frac{Y}{n} \, d\mathbb{P} \\ &\leq \frac{N}{n} \underbrace{\mathbb{P}(Y < N)}_{\leq 1} + \mathbb{P}(N \leq Y < n) \\ &\leq \frac{N}{n} + \mathbb{P}(N \leq Y < \infty) \leq \frac{N}{n} + \epsilon. \end{align*}$$
Choosing $n$ sufficiently large, we get
$$\frac{1}{n} \int_{Y<n} Y \, d\mathbb{P} \leq 2 \epsilon.$$
